On Tuesday, July 19, 2005 at 1:09:47 PM UTC+3, Robert Israel wrote:
In article <dbgogh$679$1...@dizzy.math.ohio-state.edu>,
bassam king karzeddin <bas...@ahu.edu.jo> wrote:
I will be glad to know if I wrote nonsence mathematics or something >useful.here is the problem.
An arbitrary angle and its exact trisection angle fits exactly in the >following symbolic triangle with the following sides:
a^3 , a*(b^2-a^2) , b*(b^2-2*a^2)
Where : 2 >= b/a >= sqrt(2)
(a,b):are positive real numbers
I think you mean: if theta is an angle between 0 and pi/4,
a triangle with angles theta, 3 theta and pi - 4 theta
has sides with the ratios a^3 :: a*(b^2-a^2) :: b*(b^2-2*a^2)
where cos(theta) = b/(2*a).
Yes, that's true, and elementary to verify using the Law of Sines
and addition formulas for the sine function.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Robert was proven idiot FOR SURE
BKK
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